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I am given the matrix $$B= \begin{pmatrix} 0 & 0 &0 & 1\\ 0& 0 & 1 &0 \\ 1 &0 &0 &0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$$

and I want to find its order. So, I have to find the minimum $n \in \mathbb{N}$ such that $B^n=I$.

Instead of this way,could we also calculate the number of permutations we have to do,to get the matrix $I$?

evinda
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3 Answers3

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Rather than the number of permutations, you have to compute the order of the permutation which corresponds to this matrix. The permutation in question is $$\begin{pmatrix} 1 & 2 & 3 & 4\\ 3 & 4 & 2 & 1 \end{pmatrix},$$ which is the cycle $(1324)$, which has order $4$. Hence, the $n$ in question is $4$.

Jesko Hüttenhain
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One way of doing this would be to let the matrix act on the vector $\langle 1, 2, 3, 4 \rangle$, recognizing that the matrix will simply permute the entries of this vector.

When we let the matrix act on the vector once, we get the vector $\langle 4, 3, 1, 2 \rangle$. The resulting permutation is a $4$-cycle in $S_4$, which has order $4$. Therefore, $A^4 = I$.

Side note:

In general, if you are given the set $M$ of all $n \times n$ matrices whose rows are composed of the standard basis vectors $e_1, e_2, ..., e_n$, then there is an isomorphism $\phi:M \rightarrow S_n$. See this article.

Kaj Hansen
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The characteristic polynomial is:

$$p(\lambda)=\det(B-\lambda I)=\left| \begin{matrix} -\lambda & 0 &0 & 1\\ 0& -\lambda & 1 &0 \\ 1 &0 & -\lambda &0 \\ 0 & 1 & 0 & -\lambda \end{matrix}\right|=\lambda^4-1$$

So by Cayley–Hamilton theorem:

$$p(B)=0$$ $$B^4-I=0$$ $$B^4=I$$

rlartiga
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  • As a matter of fact, the characteristic polynomial of a permutation matrix is closely linked to the cycle structure of the underlying permutation, so you don't have to compute this determinant and Jesko's answer is perfectly fine ;-) – Jean-Claude Arbaut May 25 '14 at 22:14
  • @Jean-ClaudeArbaut I don't have idea of that relationship you have something to read about it? – rlartiga May 25 '14 at 22:16
  • I'll look for a better reference, but you may have a look at this, section 2.2. I think I had an exercise on this, long ago. – Jean-Claude Arbaut May 25 '14 at 22:42
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    Couldn't the order still be any divider of 4? – Klaas van Aarsen May 25 '14 at 22:53
  • @IlikeSerena according with http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem is the minimal polynomial which satisfies $P(A)=0$ – rlartiga May 25 '14 at 23:50
  • @rlartiga: That is true for the minimal polynomial, but not necessarily for the characteristic polynomial. At this time we only know that the minimal polynomial divides the characteristic polynomial. – Klaas van Aarsen May 26 '14 at 19:14